# Activities - progressions problems

# Problem 1 from Exercise 3.2 ( Q.N.11 - page No. 37)

A company employed 400 persons in the year 2001 and each year increased by 35 persons. In which year the number of employees in the company will be 785? Solution

## Pre-requisites

Pupil know:-

- Student know meaning of progression and arithmatic progression.
- Student know the terms used in arithmatic progression problem.
- Student know the formula under arithmatic progression .
- Student know the basic operation to calculate the values.

## Interpretation of the problem

- Student should understand there are 400 persons in a company in 2001.
- Student also know that every year they taken equal number of employees - this means that it is related to arithmatic progression

## What is asked of student

In which year the number of employees is equal to 785. This means they know the first term is 400 and last term is 785. So they have to calculate the number of terms in this sequence.

## Method 1:Solution by formula method

### Assumption

- He know the general form of A.P.
- He know the formula of nth term of A.

### Algorithm

Given:Arithmatic progression

400, 435, 470,......,785.

First term = a = 400

common differnce = d = 35

Last term = Tn = 785

Number of terms = n = ?

We know the n th term of A.P. is

Tn = a + (n-1) d

785 = 400 + ( n-1) 35

785 – 400 = 35n – 35

385 + 35 = 35n

420 = 35n

n = 12

So there are 12 terms in that progression . So In 2012 the number of employees equal to 785.

## Method 2 - By Verification

### Assumption

Student know the writing progression when he know first term and common differnce.

### Algarithm

400, 435, 470, 505, 540, 575, 610, 645, 680, 715, 750, 785.

In 2001 -------> 400

In 2002 --------> 435

In 2003---------> 470

In 2004 --------> 505

In 2005 --------> 540

In 2006 --------> 575

In 2007 ----------> 610

In 2008 ----------> 645

In 2009 -----------> 680

In 2010 ----------->715

In 2011 -----------> 750

In 2012 -----------> 785

so in 2012 the number employees became 785.

# Problem 2 from Exercise 3.3 ( Q.N.12 - page No. 43)

The sum of 6 terms which form an A.P is 345. The difference between the first and last terms is 55. Find the terms.

## Pre-requisites

Pupil know:-

- What is progression ?
- What is arithmatic progression?
- Student know the general form of A.P.
- Student know the knowledge of simultaniouse equation and substitution method.

## Premisis for problem solving

- Student know to use the general form A.P.
- Solve simultaniouse equation by substitution method

## Interpretation of the problem

- There are six terms in an A.P. and their sum is given.
- Differnce between first and last terms is given.
- We find the remaining terms of A.P.

## What is asked of student

To find the remaining terms in A.P.

## Method 1 :- Solution by formula method

### Assumption

- He know the general form of A.P.
- He know the simultaniouse linear eq.

## Algorithm

The general form A.P. is a, a+d, a+2d, a+3d, ......, a+(n-1)d.

But there are only six terms are there so

a, a+d, a+2d, a+3d, a+4d, a+5d.

Sum of six terms is 345.

so a+a+d+a+2d+a+3d+a+4d+a+5d = 345

6a + 15d = 345------------>1

Differnce between last term and first term is 55.

so (a+ 5d) – (a) = 55

5d = 55

d = 11 -------------->2

Now we find the first term of A.P. by substitute in eq 1.

6a + 15d = 345

6a + 15 (11) = 345

6a + 165 = 345

6a = 345 – 165

6a = 180

a = 30

then, a = 30, d = 11,

so A.P. is 30, 41,52,63,74,85.

# Problem 3

**(Exercise 3.7 , Problem number 8, Page number 62)**
Find two numbers whose arithmetic mean exceeds their geometric mean by 2, and whose harmonic mean is one-fifth of the larger number.

## Pre-requisites

Pupil know

- The meaning of arithmetic mean , geometric mean and harmonic mean
- The formula of A.M , G.M and H.M
- Student know the basic operation to calculate the values.

## Interpretation of the problem

- Student should take a>b
- Student know that arithmetic mean is exceeds the geometric mean by 2.
- Harmonic mean is one – fifth of the larger number.

## What is asked of student

Find the two numbers 'a' and 'b' such that whose arithmetic mean is exceeds their geometric mean by 2 and harmonic mean is one – fifth of the larger number.

## Solution of the problem

Assumption :-

- He know the formula for A.M , G.M and H.M
- He know the basic operation.

## Algorithm

Let 'a' and 'b' two numbers such that a>b

Arithmetic mean formula is A.M =

Geometric mean formula is G.M =

A.M exceeds their G.M by 2

A.M = G.M +2

= + 2

a + b = 2 ( + 2 )-------> 1

H.M is one – fifth larger number . Here 'a' is larger number so,

H.M = a

= a

Here we cancel 'a' because it is in both side of the equal sign and after cancel 'a' we get,

=

Cross multiply we get,

a + b = 10b ----------------->2

Take '10b' to left hand side, we get

a + b – 10b = 0

a – 9b = 0

a = 9b

Consider equation 1

a + b = 2 ( + 2 )

Put a = 9b in the above equation,

9b + b = 2 ( + 2 )

10b = 2 (

10b = 2 ( 3b + 2 )

10b = 6b + 4

10b – 6b = 4

4b = 4

b = 1

Consider equation 2

a + b = 10b

Put b = 1

a + (1) = 10 (1)

a + 1 = 10

a = 10-1

a = 9

# Problem 4

** Exercise 3.7 , Problem number 10, Page number 62 **
If 'a' be the arithmetic mean between 'b' and 'c', and 'b' the geometric mean between 'a' and 'c', then prove that 'c' will be the harmonic mean between 'a' and 'b'.

## Pre-requisites

Pupil know

- The meaning of arithmetic mean , geometric mean and harmonic mean
- The formula of A.M , G.M and H.M
- Student know the basic operation to calculate the values.

## Interpretation of the problem

- In this problem 'a' is the arithmetic mean of 'b' and 'c'
- They also given that 'b' is the geometric mean of 'a' and 'c'

## What is asked of student

- In this problem we prove that 'c' will be the harmonic mean between 'a' and 'b'
- That is we just show that c =

## Solution of the problem

Assumption :-

- He know the formula for A.M , G.M and H.M
- He know the basic operation.
- He know substitute the values

## Algorithm

The formula for arithmetic mean between 'a' and 'b' is A.M =

The formula for geometric mean between 'a' and 'b' is G.M =

Then formula for harmonic mean between 'a' and 'c' is H.M =

Let 'a' be the arithmetic mean of 'b' and 'c'

That is a =

Re-arrange the formula,

2 =

Multiply both side by 'ab' we get,

2ab =

Cancel 'a' in Right hand side and multiply 'b' inside to the bracket we get

2ab = + bc ---------->1

Also 'b' is the geometric mean between 'a' and 'c'

That is b =

We also write this as = ac.-------->2

Now substitute thia value In equation 1,

2ab = ac + bc

Take common in right hand side ( c is common )

2ab = c(a + b)

Divide both side by (a + b),

= c

Hence 'c' is the harmonic between 'a' and 'b'.